Class D Amplifier

The Class-D Amplifier
(From the book Introduction to Electroacoustics and Audio Amplifier Design, Second Edition - Revised Printing, by W. Marshall Leach, Jr., published by Kendall/Hunt, c ° 2001.) A class-D amplifier is one in which the output transistors are operated as switches. When a transistor is off, the current through it is zero. When it is on, the voltage across it is small, ideally zero.In each case, the power dissipation is very low. This increases the efficiency, thus requiring less power from the power supply and smaller heat sinks for the amplifier. These are important advantages in portable battery-powered equipment. The “D” in class-D is sometimes said to stand for “digital.” This is not correct because the operation of the class-D amplifier is based on analog principles. Thereis no digital coding of the signal. Before the advent of the class-D amplifier, the standard classes were class-A, class-AB, class-B, and class-C. The “D” is simply the next letter in the alphabet after “C.” Indeed, the earliest work on class-D amplifiers involved vacuum tubes and can be traced to the early 1950s. Fig. 1 shows the basic simplified circuit of a class-D amplifier. We assume a bipolarpower supply so that V − = −V + . The amplifier consists of a comparator driving two MOSFET transistors which operate as switches. The comparator has two inputs. One is a triangle wave, the other is the audio signal. The frequency of the triangle wave must be much higher than that of the audio input. The voltage output of the comparator can be written vC = −V1 for vS > vT vC = +V1 for vS < vT (1)This voltage is applied to the input of a complementary common-source MOSFET output stage. Each transistor operates as a switch. For vC = −V1 , M1 is on and M2 is off. If the voltage drop 0 0 across M1 is negligible, then vO = V + . Similarly, for vC = +V1 , M2 is on, M1 is off, and vO = V − . In practice, there is a small voltage drop across the on MOSFET switch so that the peak output 0 voltageis less than the power supply voltage. For the case vS = 0, vO is a symmetrical square wave. The low-pass filter consisting of L1 and C1 passes the average value of the square wave to the loudspeaker, which is zero. Thus vO = 0 for vS = 0. The network consisting of R1 and C2 compensates for the inductive impedance of the loudspeaker voice coil so that the filter sees a resistive load at highfrequencies.

Figure 1: Basic class-D amplifier. Fig. 2 shows the circuit waveforms for the case where vS is a sine wave. For purposes of illustration, the sine wave frequency is fS = 1 kHz and the triangle wave frequency is fT = 20 kHz. The sine wave amplitude is 0.75VT P . For vS > 0, the duty cycle of the square wave changes so 0 0 that vO spends more time at its positive level than at its negativelevel. This causes vO to have a 0 0 positive average value. Similarly, for vS < 0, vO has a negative average value. The waveform for vO 1

is said to be pulse-width-modulated. The passive filter consisting of L1 and C1 passes the average 0 or low-frequency value of vO to the loudspeaker load and rejects the higher-frequency harmonics of the switching waveform.

Figure 2: Amplifier voltagewaveforms. The effective gain of the amplifier can be determined by applying a dc voltage at the input and 0 0 0 calculating the ratio of hvO i to vS , where hvO i denotes the low-frequency time average of vO . If vS 0 is increased, hvO i increases linearly until it reaches the level VOP , which corresponds to the positive clipping voltage at the output. This occurs when vS = VT P . It follows that theeffective gain k is given by hv0 i VOP k= O = (2) vS VT P Fig. 3 shows the waveforms of the output voltage vO for two values of the cutoff frequency of the LC filter. The transfer function of the filter is Vo 1 = 2 0 Vo (s/ω c ) + (1/Qc ) (s/ω c ) + 1 (3)

√ where ω c = 2πfc = 1/ L1 C1 is the resonance frequency and Qc = 1/ (ω c RL C1 ) is the quality factor. The load resistance RL is the effective...